Q.If $|\bar a|=3,|\bar b|=4$, then the value of $\lambda$ for which $\bar a+\lambda\bar b$ is perpendicular to $\bar a-\lambda\bar b$, is (A) $\frac{9}{16}$ (B) $\frac{3}{4}$ (C) $\frac{3}{2}$ (D) $\frac{4}{3}$

Perpendicular vectors have dot product zero: $\bar a\cdot\bar b=0$.

$\bar a\cdot\bar b=(3)(1)+(2)(p)+(9)(3)=3+2p+27=2p+30$

For $\bar a\perp\bar b$: $2p+30=0 \implies p=-15$.

Use $|\bar a\times\bar b|=|\bar a||\bar b|\sin\theta$ to find $\theta$, then $|\bar a\cdot\bar b|=|\bar a||\bar b|\cos\theta$.

Given $|\bar a|=5$, $|\bar b|=13$, $|\bar a\times\bar b|=25$.

$|\bar a\times\bar b|=|\bar a||\bar b|\sin\theta \implies 25 = 5\cdot13\cdot\sin\theta = 65\sin\theta$

$\sin\theta=\frac{25}{65}=\frac{5}{13}$

Since $\sin\theta=5/13$, $\cos\theta=\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13}$.

$|\bar a\cdot\bar b|=|\bar a||\bar b|\cos\theta = 5\cdot13\cdot\frac{12}{13}=5\cdot12=60$

$|\bar a+\bar b|=|\bar a-\bar b|$ forces $\bar a\cdot\bar b=0$, so the vectors are perpendicular.

Step 1 — Square both sides.

$|\bar a+\bar b|^2 = |\bar a-\bar b|^2$

$(\bar a+\bar b)\cdot(\bar a+\bar b) = (\bar a-\bar b)\cdot(\bar a-\bar b)$

$|\bar a|^2+2\bar a\cdot\bar b+|\bar b|^2 = |\bar a|^2-2\bar a\cdot\bar b+|\bar b|^2$

Step 2 — Simplify.

$4\,\bar a\cdot\bar b = 0 \implies \bar a\cdot\bar b = 0$

Step 3 — Interpret.

Since $\bar a,\bar b$ are non-zero, $\bar a\cdot\bar b = |\bar a||\bar b|\cos\theta = 0$ forces $\cos\theta=0$, i.e. $\theta = 90^\circ$.

Two vectors are perpendicular exactly when their dot product is zero; solving that equation gives $\lambda$.

Given $\bar{u} = \bar{i} - \lambda\bar{j} + 2\bar{k}$ and $\bar{v} = 8\bar{i} + 6\bar{j} - \bar{k}$.

Step 1. Vectors are at right angles iff $\bar{u}\cdot\bar{v} = 0$.

Step 2. Compute the dot product:

$\bar{u}\cdot\bar{v} = (1)(8) + (-\lambda)(6) + (2)(-1) = 8 - 6\lambda - 2 = 6 - 6\lambda$

Step 3. Set equal to zero: $6 - 6\lambda = 0 \Rightarrow \lambda = 1$.

Two vectors are perpendicular exactly when their dot product is zero — set the dot product to zero and solve the resulting quadratic in $\lambda$.

Given $\bar{u}=\lambda\bar{i}-3\bar{j}+5\bar{k}$ and $\bar{v}=2\lambda\bar{i}-\lambda\bar{j}-\bar{k}$, perpendicular means $\bar{u}\cdot\bar{v}=0$.

$\bar{u}\cdot\bar{v} = \lambda(2\lambda) + (-3)(-\lambda) + 5(-1) = 2\lambda^2 + 3\lambda - 5 = 0$

Solve using the quadratic formula:

$\lambda = \dfrac{-3 \pm \sqrt{9+40}}{4} = \dfrac{-3\pm7}{4}$

$\lambda = \dfrac{4}{4}=1$ or $\lambda = \dfrac{-10}{4}=-\dfrac{5}{2}$